complex integral, partial fractions 2











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I have to evaluate the following integral:
$ int_{gamma} frac{1}{z^3-z^2+z-1} dz$, $ gamma: [0, 2pi] rightarrow mathbb{C}, gamma(t)=1+e^{2it} $



My idea was to split the integrand into partial fractions and apply cauchys integral formula. Then i get:



$int_{gamma} frac{1}{z^3-z^2+z-1} dz = frac{1}{2}int_{gamma} frac{1}{z-1}dz- frac{1}{4} int_{gamma} frac{1-i}{z-i}dz- frac{1}{4} int_{gamma} frac{1-i}{z+i}dz $
The last 2 integrals are 0, because $ pm i $ arent enclosed by the given contour. For the first one, i get using the cauchy integral formula: $ 2 pi i cdot ind_{Gamma} cdot 2 = int... $. Therefore $int... = 2 pi i cdot 2 cdot 2 = 8 pi i$
Is that right?










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  • Is my solution right?
    – Sarah34
    Nov 25 at 9:49















up vote
1
down vote

favorite












I have to evaluate the following integral:
$ int_{gamma} frac{1}{z^3-z^2+z-1} dz$, $ gamma: [0, 2pi] rightarrow mathbb{C}, gamma(t)=1+e^{2it} $



My idea was to split the integrand into partial fractions and apply cauchys integral formula. Then i get:



$int_{gamma} frac{1}{z^3-z^2+z-1} dz = frac{1}{2}int_{gamma} frac{1}{z-1}dz- frac{1}{4} int_{gamma} frac{1-i}{z-i}dz- frac{1}{4} int_{gamma} frac{1-i}{z+i}dz $
The last 2 integrals are 0, because $ pm i $ arent enclosed by the given contour. For the first one, i get using the cauchy integral formula: $ 2 pi i cdot ind_{Gamma} cdot 2 = int... $. Therefore $int... = 2 pi i cdot 2 cdot 2 = 8 pi i$
Is that right?










share|cite|improve this question






















  • Is my solution right?
    – Sarah34
    Nov 25 at 9:49













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have to evaluate the following integral:
$ int_{gamma} frac{1}{z^3-z^2+z-1} dz$, $ gamma: [0, 2pi] rightarrow mathbb{C}, gamma(t)=1+e^{2it} $



My idea was to split the integrand into partial fractions and apply cauchys integral formula. Then i get:



$int_{gamma} frac{1}{z^3-z^2+z-1} dz = frac{1}{2}int_{gamma} frac{1}{z-1}dz- frac{1}{4} int_{gamma} frac{1-i}{z-i}dz- frac{1}{4} int_{gamma} frac{1-i}{z+i}dz $
The last 2 integrals are 0, because $ pm i $ arent enclosed by the given contour. For the first one, i get using the cauchy integral formula: $ 2 pi i cdot ind_{Gamma} cdot 2 = int... $. Therefore $int... = 2 pi i cdot 2 cdot 2 = 8 pi i$
Is that right?










share|cite|improve this question













I have to evaluate the following integral:
$ int_{gamma} frac{1}{z^3-z^2+z-1} dz$, $ gamma: [0, 2pi] rightarrow mathbb{C}, gamma(t)=1+e^{2it} $



My idea was to split the integrand into partial fractions and apply cauchys integral formula. Then i get:



$int_{gamma} frac{1}{z^3-z^2+z-1} dz = frac{1}{2}int_{gamma} frac{1}{z-1}dz- frac{1}{4} int_{gamma} frac{1-i}{z-i}dz- frac{1}{4} int_{gamma} frac{1-i}{z+i}dz $
The last 2 integrals are 0, because $ pm i $ arent enclosed by the given contour. For the first one, i get using the cauchy integral formula: $ 2 pi i cdot ind_{Gamma} cdot 2 = int... $. Therefore $int... = 2 pi i cdot 2 cdot 2 = 8 pi i$
Is that right?







integration






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asked Nov 25 at 0:06









Sarah34

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  • Is my solution right?
    – Sarah34
    Nov 25 at 9:49


















  • Is my solution right?
    – Sarah34
    Nov 25 at 9:49
















Is my solution right?
– Sarah34
Nov 25 at 9:49




Is my solution right?
– Sarah34
Nov 25 at 9:49















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